Question

Let $f\left(x\right)={\int }_0^{1}t|t-x|dt,xϵR$
The minimum value of f(x) is

• A. $\frac{1}{3}(1-\frac{1}{\sqrt{2}})$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{3}-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $\frac{1}{3}(1-\frac{1}{\sqrt{2}})$

${\int }_0^{1}t|t-x|dx$
$x<0\Rightarrow f\left(x\right){\int }_0^{1}t(t-x)dx={(\frac{t^3}{3}-\frac{t^2}{2}x)}_0^1=\frac{1}{3}-\frac{x}{2}$
$x>0\Rightarrow f\left(x\right){\int }_0^{1}t(x-t)dt={(\frac{t^{2}x}{2}-\frac{t^2}{3})}_0^1=\frac{x}{2}-\frac{1}{3}$
$0\le x\le 1={\int }_0^x(x-t)dt+{\int }_0^{x}t(x-t)dt$
${(\frac{t^{2}x}{2}-\frac{t^3}{3})}_0^x+{(\frac{t^3}{3}-\frac{x{t}^2}{2})}_x^1$
$=(\frac{x^2}{2}-\frac{x^3}{3})+(\frac{1}{3}-\frac{x}{2})-(\frac{x^3}{3}-\frac{x^3}{2})$
$f\left(x\right)=\frac{x^3}{2}-\frac{x}{2}+\frac{1}{3}$
Minimum value is possible at $x=\frac{1}{\sqrt{2}}$
$f\left(6\right)=\frac{8}{3}$